3.2594 \(\int \frac {(3+5 x)^{3/2}}{(1-2 x)^{5/2} (2+3 x)} \, dx\)

Optimal. Leaf size=79 \[ \frac {2 (5 x+3)^{3/2}}{21 (1-2 x)^{3/2}}-\frac {2 \sqrt {5 x+3}}{49 \sqrt {1-2 x}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{49 \sqrt {7}} \]

[Out]

2/21*(3+5*x)^(3/2)/(1-2*x)^(3/2)-2/343*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-2/49*(3+5*x)^(1
/2)/(1-2*x)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {94, 93, 204} \[ \frac {2 (5 x+3)^{3/2}}{21 (1-2 x)^{3/2}}-\frac {2 \sqrt {5 x+3}}{49 \sqrt {1-2 x}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{49 \sqrt {7}} \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*x)^(3/2)/((1 - 2*x)^(5/2)*(2 + 3*x)),x]

[Out]

(-2*Sqrt[3 + 5*x])/(49*Sqrt[1 - 2*x]) + (2*(3 + 5*x)^(3/2))/(21*(1 - 2*x)^(3/2)) - (2*ArcTan[Sqrt[1 - 2*x]/(Sq
rt[7]*Sqrt[3 + 5*x])])/(49*Sqrt[7])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin {align*} \int \frac {(3+5 x)^{3/2}}{(1-2 x)^{5/2} (2+3 x)} \, dx &=\frac {2 (3+5 x)^{3/2}}{21 (1-2 x)^{3/2}}-\frac {1}{7} \int \frac {\sqrt {3+5 x}}{(1-2 x)^{3/2} (2+3 x)} \, dx\\ &=-\frac {2 \sqrt {3+5 x}}{49 \sqrt {1-2 x}}+\frac {2 (3+5 x)^{3/2}}{21 (1-2 x)^{3/2}}+\frac {1}{49} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {2 \sqrt {3+5 x}}{49 \sqrt {1-2 x}}+\frac {2 (3+5 x)^{3/2}}{21 (1-2 x)^{3/2}}+\frac {2}{49} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )\\ &=-\frac {2 \sqrt {3+5 x}}{49 \sqrt {1-2 x}}+\frac {2 (3+5 x)^{3/2}}{21 (1-2 x)^{3/2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{49 \sqrt {7}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 71, normalized size = 0.90 \[ \frac {2 \left (7 \sqrt {5 x+3} (41 x+18)+3 \sqrt {7-14 x} (2 x-1) \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )\right )}{1029 (1-2 x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*x)^(3/2)/((1 - 2*x)^(5/2)*(2 + 3*x)),x]

[Out]

(2*(7*Sqrt[3 + 5*x]*(18 + 41*x) + 3*Sqrt[7 - 14*x]*(-1 + 2*x)*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])]))/
(1029*(1 - 2*x)^(3/2))

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fricas [A]  time = 0.77, size = 86, normalized size = 1.09 \[ -\frac {3 \, \sqrt {7} {\left (4 \, x^{2} - 4 \, x + 1\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (41 \, x + 18\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{1029 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+5*x)^(3/2)/(1-2*x)^(5/2)/(2+3*x),x, algorithm="fricas")

[Out]

-1/1029*(3*sqrt(7)*(4*x^2 - 4*x + 1)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x
- 3)) - 14*(41*x + 18)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)

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giac [A]  time = 1.13, size = 113, normalized size = 1.43 \[ \frac {1}{3430} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {2 \, {\left (41 \, \sqrt {5} {\left (5 \, x + 3\right )} - 33 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{3675 \, {\left (2 \, x - 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+5*x)^(3/2)/(1-2*x)^(5/2)/(2+3*x),x, algorithm="giac")

[Out]

1/3430*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/
(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) + 2/3675*(41*sqrt(5)*(5*x + 3) - 33*sqrt(5))*sqrt(5*x +
3)*sqrt(-10*x + 5)/(2*x - 1)^2

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maple [B]  time = 0.01, size = 154, normalized size = 1.95 \[ \frac {\left (12 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-12 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+574 \sqrt {-10 x^{2}-x +3}\, x +3 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+252 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}\, \sqrt {5 x +3}}{1029 \left (2 x -1\right )^{2} \sqrt {-10 x^{2}-x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x+3)^(3/2)/(-2*x+1)^(5/2)/(3*x+2),x)

[Out]

1/1029*(12*7^(1/2)*x^2*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-12*7^(1/2)*x*arctan(1/14*(37*x+20)*7
^(1/2)/(-10*x^2-x+3)^(1/2))+3*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+574*(-10*x^2-x+3)^(1/
2)*x+252*(-10*x^2-x+3)^(1/2))*(-2*x+1)^(1/2)*(5*x+3)^(1/2)/(2*x-1)^2/(-10*x^2-x+3)^(1/2)

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maxima [A]  time = 1.21, size = 104, normalized size = 1.32 \[ \frac {1}{343} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {205 \, x}{147 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {125 \, x^{2}}{6 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {37}{588 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {1385 \, x}{84 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {67}{28 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+5*x)^(3/2)/(1-2*x)^(5/2)/(2+3*x),x, algorithm="maxima")

[Out]

1/343*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) - 205/147*x/sqrt(-10*x^2 - x + 3) + 125/6*x^2/
(-10*x^2 - x + 3)^(3/2) - 37/588/sqrt(-10*x^2 - x + 3) + 1385/84*x/(-10*x^2 - x + 3)^(3/2) + 67/28/(-10*x^2 -
x + 3)^(3/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (5\,x+3\right )}^{3/2}}{{\left (1-2\,x\right )}^{5/2}\,\left (3\,x+2\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + 3)^(3/2)/((1 - 2*x)^(5/2)*(3*x + 2)),x)

[Out]

int((5*x + 3)^(3/2)/((1 - 2*x)^(5/2)*(3*x + 2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+5*x)**(3/2)/(1-2*x)**(5/2)/(2+3*x),x)

[Out]

Timed out

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